Module P2.2 Projectile motion

Transcription

1 F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S Module P2.2 Projectile motion 1 Opening items 1.1 Module introduction 1.2 Fast track questions 1.3 Ready to study? 2 Describing the motion of a projectile 2.1 Position and displacement 2.2 Vector algebra 2.3 Velocity in projectile motion 2.4 Acceleration in projectile motion 2.5 The independence of x- and y-motions for projectiles 3 Applying the equations of motion 3.1 Horizontal motion 3.2 Vertical motion 3.3 The trajectory of a projectile 3.4 The range of a projectile 4 Solving projectile problems 4.1 Some examples 4.2 The vector equations for motion with uniform acceleration 4.3 Solution to the introductory problem 5 Closing items 5.1 Module summary 5.2 Achievements 5.3 Exit test Exit module

2 1 Opening items 1.1 Module introduction Any object that is launched into unpowered flight close to the Earth s surface is called a projectile. Examples include artillery shells, shot putts, high jumpers and balls in games such as tennis, football and cricket. To a good approximation (ignoring any sideways swerve) many projectiles can be modelled as particles moving along curved paths in a vertical plane. The motion is therefore two-dimensional and the following questions can be asked: What is the horizontal range of the projectile and what is the maximum height that it attains? How long does the flight take, and what is the shape of the flight path? How do the velocity and acceleration of the projectile vary during its flight? This module explains how questions such as these can be answered. In doing so it also provides an introduction to the general analysis of two-dimensional motion in terms of vectors and their (scalar) components. In particular, Section 2 explains how vectors can be used to describe quantities such as the position, displacement, velocity and acceleration of a projectile, and how vectors of a similar type can be added together and algebraically manipulated.

3 Section 3 investigates the equations that describe the motion of a projectile, stressing the fact that the horizontal and vertical motions of a projectile are essentially independent, apart from having the same duration. The equations that emerge from this investigation are used to deduce a number of general features of projectile motion, including the shape of the projectile s trajectory and the condition for achieving the maximum horizontal range. Finally, in Section 4, the techniques developed earlier are used to solve a variety of twodimensional projectile problems, and their extension to three-dimensional problems involving arbitrary uniform acceleration is briefly mentioned. The following problem will give you an idea of the sort of question you will be able to answer by the end of this module. The solution is given in Subsection 4.3. An aircraft flies at a height of 20001m with a constant velocity of 1501m1s 1 in a straight horizontal line. As it passes vertically over a gun a shell is fired from the gun. Find the minimum muzzle speed, u, of the shell, and the angle φ, from the horizontal, at which the shell should be fired in order for it to hit the plane. Take the magnitude of the acceleration due to gravity g as 9.811m1s 2 and ignore air resistance. Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the Fast track questions given in Subsection 1.2. If not, proceed directly to Ready to study? in Subsection 1.3.

4 1.2 Fast track questions Study comment Can you answer the following Fast track questions?. If you answer the questions successfully you need only glance through the module before looking at the Module summary (Subsection 5.1) and the Achievements listed in Subsection 5.2. If you are sure that you can meet each of these achievements, try the Exit test in Subsection 5.3. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module. Question F1 Miss Magnificent, the human cannonball, is fired from her cannon with a muzzle speed of 181m1s 1 at an angle of 35 to the horizontal. If she is to land at the same level from which she took off, how far away from the cannon should the net in which she wishes to land be placed? You should ignore the effect of air resistance and treat Miss Magnificent as a point particle.

5 Question F2 A particle moves in the (x, y) plane from point 1, with position (31m, 21m) to point 2, with position (71m, 11m). (a) What are the x- and y-components of the particle s displacement from point 1 to point 2? (b) What is the magnitude of this displacement? (c) What is the angle between this displacement and the x-axis? Question F3 A shell is fired with a velocity of 2001m1s 1 at an angle of 30 above the horizontal. Find the time taken for the shell to reach its maximum height and the magnitude and direction of its velocity after 16 s. Take the magnitude of the acceleration due to gravity to be g = 9.811m1s 2 and ignore the effect of air resistance.

6 Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to Ready to study? in Subsection 1.3. Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Closing items.

7 1.3 Ready to study? Study comment To begin to study of this module you will need to be familiar with the following terms or topics: Cartesian coordinate system, cosine (as in cos1θ), gradient of a line, graph, position coordinates, Pythagoras s theorem, quadratic equation, SI units, sine (as in sin1θ) and tangent (as in tan1θ). It is also assumed that you have some familiarity with concepts such as position, speed, velocity and acceleration in the context of one-dimensional linear motion and that you have previously encountered the calculus notation dx/dt used to indicate the (instantaneous) rate of change of x with respect to t. However, proficiency in the use of calculus is not required for the study of this module. If you are uncertain about any of these items you can review them now by reference to the Glossary, which will also indicate where in FLAP they are developed. The following Ready to study questions will enable you to check whether you need to review some of the topics before embarking on this module.

8 Question R1 A right-angled triangle has sides of length a, b and c. If the longest side is of length c and if the angle between that side and side b is θ, write down Pythagoras s theorem as it applies to this particular triangle and express the lengths a and b in terms of c and θ. Question R2 An object moving along the x-axis of a Cartesian coordinate system does so in such a way that its position coordinate x changes from x = 1.01m to x = 8.01m in a time interval of 3.01s. What is the average velocity 1v x 1 of the object over the three-second interval? In what way would your answer have been different if the initial and final values of x had been interchanged? Question R3 A cyclist is travelling due north along a straight road at 101m1s 1 when the brakes are applied. If the brakes can cause the speed to reduce at a constant rate of 31m1s 2, how long will it take to stop the bicycle and how far will it have travelled in that time?

9 2 Describing the motion of a projectile Any object that is launched into unpowered flight near the Earth s surface is called a projectile. A rock thrown from a hand or an arrow released from a bow would be typical examples. The motion of a projectile has been of interest since ancient times, but describing the motion accurately and explaining its cause has created a great deal of difficulty. In the 4th century BC the Greek philosopher Aristotle ( BC) said that when an object is thrown it will carry on in a straight line until it runs out of force, after which it will fall down. Figure 1 illustrates his idea, which we will soon see to be incorrect. Figure 14Aristotle s idea of projectile motion. The dots represent successive positions of the projectile at equally separated moments of time.

10 In medieval times the exact paths of cannonballs and arrows were of great interest, and it became necessary to think about them more precisely. In the 16th century the Italian scientist Galileo Galilei ( ) produced an accurate analysis of the flight path, or trajectory, of a projectile, this is illustrated schematically in Figure 2. Figure 24Galileo s idea of projectile motion. Note how the dots representing successive positions of the projectile bunch together near the top of the trajectory.

11 In this module we will be concerned only with projectiles that may be treated as particles (i.e. bodies with negligible size and no internal structure, that may be regarded as occupying a single point in space at any particular time). Also we will assume that during flight, projectiles move under the influence of gravity alone; in other words we will ignore air resistance and other such effects. These restrictions may seem rather limiting, but for many purposes the motion of quite large bodies, such as footballs and people, can often be modelled quite successfully on this basis. Of course, there are many problems in which these simplifications would be misleading. Real projectiles have a finite size and this allows air resistance to influence their motion. When air resistance is large it cannot be ignored, for example, badminton players will recognize the trajectory of a shuttlecock as being nearer to that shown in Figure 1 than Figure 2. Also, in many ball games, such as cricket and tennis, it is possible to spin the ball. The effect of this spin on the flow of air around the ball produces a sideways force so that the ball swerves. These phenomena are beyond the scope of this module, though it is not too difficult to extend the ideas that are introduced here in order to accommodate them.

12 2.1 Position and displacement Study comment This subsection is mainly concerned with vector notation and deliberately parallels the development of vectors presented in the maths strand of FLAP. If you have already studied vectors there or elsewhere you should aim to finish this subsection very quickly, pausing only to make sure that you are familiar with the notation that will be used in this module and that you are able to answer Questions T1 and T2. If you are unfamiliar with vectors and you feel you need a more detailed treatment than that presented here you should consult the Glossary which will refer you to the appropriate modules in the maths strand. Position A projectile represented by a point particle moving under the influence of gravity has a trajectory that is confined to a single vertical plane. Such a projectile is said to exhibit two-dimensional motion, since its trajectory may be adequately described using just two independent coordinate axes. In this module we will use two mutuallyperpendicular Cartesian coordinate axes for this purpose: a horizontal axis which will usually be labelled x, and a vertical axis which will usually be labelled y. These axes meet at a point called the origin from which we can measure the position coordinates of any point in the plane of the axes (i.e. the specific values of x and y that determine the location of that point in the plane). So, provided we choose the directions of the x- and y-axes appropriately, we can suppose that the motion of any projectile we wish to consider is confined to the (x, y) plane and may be described in terms of the changing values of the projectile s x- and y-coordinates.

13 Figure 3 uses a two-dimensional Cartesian coordinate system of the sort just described to show various points along the trajectory of a projectile launched from the origin (the point marked O). As usual, the dots represent successive positions of the projectile at equally separated moments of time. The location of any point on the trajectory, such as the point marked A, can be described in terms of its position coordinates. In this particular case we can call them x A and y A since they relate to the point A, and we can adopt the convention of writing them as an ordered pair (x A, y A ), in which the first value is always the x-coordinate and the second the y-coordinate. What are the position coordinates (x A, y A ) of point A in Figure 3? y/m θ r x/m Figure 34The arrow drawn from the origin to a point such as A represents the position vector r of that point. A

14 6 5 A 4 y/m Figure 3 also indicates an alternative way of specifying the location of a point in the (x, y) plane. An arrow of given length and given orientation, with its tail at the origin, will inevitably have its tip at some specific point. Such an arrow is called the position vector of the point and is generally denoted by the bold symbol r. The bold symbol is used to distinguish the position vector (which has both magnitude and direction) from its magnitude r which is the distance of the point from the origin of coordinates. A distance r doesn t point in any particular direction and does not specify a point; a position vector r has direction and does specify a point. 3 r 2 1 θ x/m Figure 34The arrow drawn from the origin to a point such as A represents the position vector r of that point. FLAP P2.2 Projectile motion COPYRIGHT 1998 THE OPEN UNIVERSITY S570 V1.1

15 Note that in order to specify the position vector r of a particular point we must clearly define two things: 1 the magnitude of r, which is the distance r from the origin to the point in question; 2 the direction of r, which in two dimensions can be described by the angle θ measured in an anticlockwise direction from the positive x-axis to r. Using a ruler and a protractor determine the magnitude and direction of the position vector r of point A. y/m 6 5 A 4 3 r 2 1 θ x/m Figure 34The arrow drawn from the origin to a point such as A represents the position vector r of that point.

16 Aside As has already been stressed, position vectors and their magnitudes may both be represented by the same letter, but the vector is distinguished by the use of a bold typeface. This is fine in print but not of much help in work that you have to write by hand. In order to show that a handwritten letter is supposed to represent a vector you should generally put a wavy underline beneath it as in ~ a. This sort of underline is used to indicate to a printer that whatever has been underlined should be set in bold type.

17 So, we now have two different ways of specifying the location of any point in the (x, y) plane. We can either give the coordinates of the point (x, y) or we can specify the point s position vector r by giving its magnitude and direction. There is clearly an intimate relationship between these two descriptions. Indeed, it is natural to use the ordered pair of position coordinates (x, y) as a way of identifying the position vector r by writing r = (x, y) (1) When represented in this way we say that x and y are the components of the position vector r. Thus, we may say that the point A in Figure 3 has the position vector, r = (7.01m, 4.21m) or, equivalently, we may say that the x-component of r is 7.01m, and that the y-component of r is 4.21m. y/m θ r x/m Figure 34The arrow drawn from the origin to a point such as A represents the position vector r of that point. A

18 For a two-dimensional position vector, the magnitude r is related to the components x and y by Pythagoras s theorem: r = x 2 + y 2 (2) where x 2 + y 2 indicates the positive square root of x 2 + y 2. Note that this relation ensures that the magnitude of the position vector will be a positive quantity as any length must be. It is often useful to emphasize this by using the symbol 1r1 to represent the magnitude of r. You might forget that r represents a positive quantity, but you are unlikely to forget that 1r1 must be positive. The direction of a position vector may also be related to its components. For the two-dimensional position vector r = (x, y), the direction of which is specified by the angle θ, the relationship takes the form tan θ = y x (3) It also follows from basic trigonometry that the horizontal and vertical components of a position vector r are x = r1cos1θ 4and4y = r1sin1θ (4) so r = (r1cos1θ, r1sin1θ) (5)

19 Question T1 Figure 4 shows a point P and its position vector, r. Write down the position coordinates of the point, P, and use those coordinates to determine the magnitude and direction of the position vector r.4 Position vectors are not the only quantities with magnitude and direction that are of interest to us in the study of projectile motion, or indeed in physics generally. In fact, any quantity that requires both a magnitude and a direction for its complete specification is called a vector, and common examples that you will meet in this module include velocity and acceleration. As you will see later, any vector may be expressed in terms of its components though in most cases the components will not have such a simple interpretation as the components of a position vector. y/m r x/m Figure 44A point P with its position vector, r. Many physical quantities, such as mass, length, time, speed and temperature, are specified by a magnitude alone. Such quantities are collectively called scalars. Although scalars are quite distinct from vectors it should be noted that the components of a vector are actually scalar quantities (they are sometimes formally referred to as scalar components for this reason). P

20 Thus, although it would not make sense to equate a vector (which involves direction) with a scalar (which does not), it is possible to identify a vector with an ordered arrangement of scalars such as (7.01m, 4.21m) since it is the agreed ordering and the predefined coordinate system that supplies the directional information. A vector is a quantity that is characterized by a magnitude and a direction. By convention, bold typeface symbols such as r are used to represent vector quantities. The magnitude of such a vector is represented by 1r1 or r and is a non-negative scalar quantity. In diagrams, vectors are represented by arrows or directed line segments. In handwritten work vectors are distinguished by a wavy underline ( ~ a ) and the magnitude of a vector is denoted ~ a.

21 Displacement In many projectile problems a vector quantity that is of particular interest is displacement which can be used to describe a change or a difference in position. The distinction between the position and displacement vectors can be seen from Figure 5 in which O is the origin of a Cartesian coordinate system from which the changing position of a projectile can be measured. During part of its motion the projectile moves from P 1 (with position vector r 1 = (x 1, y 1 )) to P 2 (with position vector r 2 = (x 2, y 2 )). The corresponding change (or difference) in the projectile s position can be represented by the arrow marked s in Figure 5. This is certainly a vector quantity since it has both magnitude and direction, but it cannot be a position vector because its tail is not at the origin. It is in fact the displacement from P 1 to P 2. y/m r 1 r 2 P 1 (x 1, y 1 ) s P 2 (x 2, y 2 ) x/m Figure 54Displacement and position vectors.

22 Like all vectors, the displacement s from P 1 to P 2 can be expressed in terms of its components. These are generally denoted by s x and s y ; so we can write s = (s x, s y ), where s x and s y, respectively, represent the differences in the x- and y-coordinates of P 1 and P 2 : s x = (x-coordinate of P 2 ) (x-coordinate of P 1 ) = x 2 x 1 s y = (y-coordinate of P 2 ) (y-coordinate of P 1 ) y/m r 1 r 2 P 1 (x 1, y 1 ) s P 2 (x 2, y 2 ) = y 2 y 1 Thus, s = (s x, s y ) = (x 2 x 1, y 2 y 1 ) (6) x/m Figure 54Displacement and position vectors. What are the components of the particular displacement shown in Figure 5? Use your answers to write the vector as an ordered pair of components.

23 The great advantage of using displacements rather than position vectors when describing projectile motion is that the displacement from one point to another does not depend in any way on where the origin of the coordinate system is located. For instance, even if we moved the origin of coordinates in Figure 5 from the origin to some other point, it would still remain true that the displacement from P 1 to P 2 is s = (2.01m, 2.41m); choosing a new origin would change the position coordinates and position vectors of P 1 and P 2, but it would not change the differences that determine s x and s y. Why is this an advantage? In projectile problems we are usually interested in how far y/m r 1 P 1 (x 1, y 1 ) x/m Figure 54Displacement and position vectors. r 2 s P 2 (x 2, y 2 ) the projectile has travelled (horizontally or vertically) from its launch point. If we describe the motion in terms of displacements from the launch point rather than positions relative to the origin we will have a description (in terms of displacements) that will be equally valid whether or not the launch point happens to be at the origin.

24 In Figure 5 the launch point was at the origin, but that won t always be the case. Although we have gone to some lengths to stress the differences between position vectors and displacements it is important to realize there are also deep similarities. A displacement vector provides us with a way of describing the location of one point (such as P 2 ) with respect to some arbitrarily chosen reference point (such as P 1 ). A position vector provides us with a way of describing the location of one point (such as P 2 ) with respect to a very particular reference point1 1the origin O. Clearly, if we were to choose O as the arbitrary reference point from which we measured displacements then the displacement of a point such as P 2 would be y/m r 1 P 1 (x 1, y 1 ) x/m Figure 54Displacement and position vectors. r 2 s P 2 (x 2, y 2 ) equal to its position vector. To this extent position vectors are simply a special class of displacements; position vectors are displacements from the origin.

25 2.2 Vector algebra It is a general property of displacements that they can be added together, though the word added has to be interpreted in a rather special way. For example, the projectile launched from point O in Figure 5 will at some stage arrive at point P 1, where its displacement from O is r 1, and some time later, after undergoing a further displacement s from P 1, it will arrive at P 2 where its displacement from O is r 2. Consequently it makes sense to say that the result of adding the displacement s to the displacement r 1, is the displacement r 2. More formally we say that r 2 is the vector sum or resultant of r 1 and s and we write r 2 = r 1 + s y/m r 1 r 2 P 1 (x 1, y 1 ) s P 2 (x 2, y 2 ) x/m Figure 54Displacement and position vectors.

26 The operation of adding vectors together is called vector addition, the process is not restricted to displacements but it can only be applied to vectors of similar type, e.g. we can add a velocity to a velocity, or an acceleration to an acceleration, but we cannot add a velocity to an acceleration. A vector sum such as r 1 + s looks a lot like an ordinary (scalar) sum, but it is really quite different because the directions of the vectors must be taken into account as well as their magnitudes. When thinking about vector addition you should have in mind a picture something like that shown in Figure 6, the essential features of which are described by the following rule: The triangle rule for adding vectors: Let a and b be vectors of similar type represented by appropriate arrows (or directed line segments). If the arrow representing b is drawn from the head of the arrow representing a, then an arrow from the tail of a to the head of b represents the vector sum c = a + b. y a c c = a + b Figure 64The triangle rule for adding vectors. Note the directions of the arrows: the diagram would be incorrect if any one, or any two of the arrows were reversed. b x If c = a + b, will it necessarily be the case that 1c1 = 1a1 + 1b1?

27 Although it is helpful to have a picture in mind when thinking about vector addition, it is usually easier to perform such additions with the aid of components. For instance, looking at Figure 6 you should be able to convince yourself that if a = (a x, a y ), b = (b x, b y ) and c = (c x, c y ) then the vector equation a + b = c may be written in the form (a x, a y ) + (b x, b y ) = (c x, c y ) where c x = a x + b x and c y = a y + b y In other words, To add together vectors a and b, simply add the corresponding components so that: a + b = (a x, a y ) + (b x, b y ) = (a x + b x, a y + b y ) (7) y a c c = a + b Figure 64The triangle rule for adding vectors. Note the directions of the arrows: the diagram would be incorrect if any one, or any two of the arrows were reversed. b x If c = a + b, where a = (1.61m, 5.51m) and b = ( 1.61m, 5.31m), express c in terms of its components and describe its direction in words.

28 Apart from adding vectors another useful mathematical operation that can be carried out is that of multiplying a vector by a scalar. This is called scaling and may be defined in the following way To multiply a vector a by a scalar λ, simply multiply each component of a by λ so that: λa = λ(a x, a y ) = (λa x, λa y ) (8) The result of scaling a by λ is to produce a vector λa of magnitude 1λa1 = 1λ1 1 1a1 that points in the same direction as a if λ is positive, and in the opposite direction to a if λ is negative. So, for example, given a displacement a, the scaled vector 2a would be twice as long and would point in the same direction, and the scaled vector a = ( 1)a would have the same length as a but would point in the opposite direction (i.e. it would be antiparallel) to a. If a = ( 71m, 41m) and b = ( 11m, 21m) what are the components of the vector a 4b? What is the magnitude of a 4b?

29 Now that you know how to add and scale vectors you should be able to rearrange vector equations. For instance, referring back to Figure 5 and recalling that r 1 = (x 1, y 1 ), r 2 = (x 2, y 2 ) and s = (x 2 x 1, y 2 y 1 ), it is easy to see that the vector equation r 2 = r 1 + s can be rearranged to give a purely vectorial definition of displacement: s = r 2 r 1 Given the rule for adding vectors, it should be pretty clear what we mean by a vector difference such as r 2 r 1 but if you are in any doubt just regard the vector difference as the sum of r 2 and the scaled vector ( 1)r 1. It then follows from the above definitions that y/m r 1 r 2 P 1 (x 1, y 1 ) s P 2 (x 2, y 2 ) x/m Figure 54Displacement and position vectors. s = r 2 r 1 = (x 2, y 2 ) (x 1, y 1 ) = (x 2 x 1, y 2 y 1 ) (9)

30 As you can see, vector equations can be treated very much like ordinary equations as far as rearrangements are concerned, though you should keep in mind the directional nature of vectors so that you avoid trying to do something silly like dividing by a vector. Another possible rearrangement of r 2 = r 1 + s is r 2 r 1 s = 0 where 0 = (0, 0) represents the zero vector which has zero magnitude. (Note that a vector cannot be equal to a scalar, so you should always write a a = 0 rather than a a = 0.) Question T2 A body moves from a point A, with position coordinates (41m, 21m), to a point B, with position coordinates (71m, 3 m). What is the magnitude and direction of this displacement? Check your answers by drawing a scale diagram.4

31 2.3 Velocity in projectile motion Let us return yet again to the kind of projectile motion shown in Figure 5, but let us now associate values of the time with various points in the motion. In particular suppose that at time t the projectile passes through a point P with position vector r = (x, y), and that a short time later, at t + t, the projectile arrives at a point with position coordinates (x + x, y + y). We can then represent the change in position over the short time t by the displacement vector r = ( x, y) and we can define the average velocity 1v1 of the projectile as it moves between the two points by v = r t = 1 ( x, y)= x t t, y t y/m r 1 r 2 P 1 (x 1, y 1 ) s P 2 (x 2, y 2 ) x/m Figure 54Displacement and position vectors. (10)

32 Note that the operation of dividing the two-dimensional displacement r by t really amounts to scaling r by 1/ t, so the resulting average velocity is another two-dimensional vector that points in the same direction as r. The x-component of 1v1 is given by x/ t and represents the average rate of change of the x-coordinate of the projectile, while the y-component, y/ t, represents the average rate of change of the projectile s y-coordinate. This definition of the average velocity vector provides the natural two-dimensional generalization of the definition given elsewhere in FLAP for average velocity in one-dimensional (linear) motion. In effect the twodimensional velocity can be regarded as the result of two independent linear velocities, one in the x-direction and the other in the y-direction. In practice we often need to know the velocity of a projectile at a particular instant, rather than the average velocity over some specified interval. This instantaneous velocity v is also a vector quantity, the components of which, v x and v y, can be found by considering the corresponding components of average velocities taken over smaller and smaller intervals around the time in question. In mathematical terms we can indicate that the instantaneous velocity is a limiting case of the average velocity as the time interval t becomes vanishingly small by writing v = lim r t 0 t = lim x t 0 t, y t = lim x t 0 t, lim y t 0 t (11)

33 Now, if you are familiar with calculus you will know that limits of this kind are represented symbolically by derivatives, so we have v = dr dt = dx dt, dy dt (12) and since we can write v = (v x, v y ) we can see that v x = dx dt 4and4 v y = dy dt Whether you are familiar with calculus or not, you should be able to appreciate that these final equations have a simple graphical interpretation. If you were to draw a graph showing how the x-coordinate of the projectile changed with time (i.e. an x against t graph) then the gradient (i.e. slope) of that graph at any particular value of t would represent the instantaneous velocity component v x at that time. Similarly, the gradient of a y against t graph at any particular value of t would represent the value of v y at that time. Thus, each component of the projectile s instantaneous velocity represents the instantaneous rate of change of the corresponding component of the projectile s position vector.

34 It follows that the instantaneous velocity itself is the instantaneous rate of change of the position vector and at any particular time it is tangential to the trajectory (see Figure 7). In discussing the velocity of a projectile we have, so far, emphasized the use of components but, as you know, a vector can also be characterized by its magnitude and direction. The magnitude of the (instantaneous) velocity v is called the (instantaneous) speed and may be represented by v or 1v1 : it is of course a scalar quantity and it can never be negative. Thus, it makes perfectly good sense to say that the velocity component v y = 2.41m1s 1, but it would be utter nonsense to say that the speed v had the same value. For the two-dimensional motion of a projectile, the direction of the velocity v can be y/m v y 0 φ v x v x/m Figure 74The instantaneous velocity v = (v x, v y ) of a projectile is tangential to the trajectory at any point. described by the angle φ (measured in the anticlockwise direction) between the positive x-axis and the velocity vector (see Figure 7).

35 So, given a velocity v we can write down the following relationships v = (v x, v y ) with v = v = v x2 +v y2 4and4 tan φ = v y v x or v x = v1cos1φ 4and4v y = v1sin1φ Study comment Notice that in Figure 7 we have used φ to represent the angle between v and the x-axis, whereas in Figure 3 we used θ to represent the angle between r and the x-axis. In problems involving projectile motion we must be careful never to confuse these two angles, which are usually quite different. In this module we will reinforce the distinction by always using the appropriate symbol for each angle. Question T3 Find the magnitude and direction of a velocity which has an x-component, v x, of 81m1s 1 and which has a y- component, v y, of 101m1s 1.4

36 Figure 8 is an enlarged version of Figure 3, without the position vector. We can use this diagram to investigate the motion of the projectile in the x- and y-directions. The dots represent successive positions of the projectile at intervals of 0.11s. Question T4 Using Figure 8, measure the x-component of the displacement between the first and second, second and third, eighth and ninth, fourteenth and fifteenth, and sixteenth and seventeenth dots. What do your answers suggest about the x- component of the projectile s velocity?4 y/m x/m Figure 84Successive positions of a projectile at intervals of 0.11s. It seems from the solution to Question T4 that the horizontal component of the velocity of a projectile is constant.

37 We can also use Figure 8 to try to acquire information about the vertical component of the velocity of a projectile. Question T5 Using Figure 8, measure the y-component of displacement between the same points as in Question T4. What do these values suggest about the y-component of the projectile s velocity?4 y/m From the solution to Question T5, the vertical component of the velocity, v y, is not constant. Its value is positive at the launch point but decreases as the projectile ascends, v y = 0 momentarily when the projectile reaches its x/m Figure 84Successive positions of a projectile at intervals of 0.11s. maximum altitude and thereafter v y is negative and decreasing until the projectile hits the ground. Clearly, if we want to describe the velocity of the projectile more precisely we need some way of describing the rate of change of this vertical component of velocity; that is the function of the next subsection.

38 2.4 Acceleration in projectile motion Just as we can use a two-dimensional velocity to describe the rate of change of position of the projectile so we can use a two-dimensional instantaneous acceleration to describe the rate of change of the projectile s velocity. So, if the velocity of the projectile changes from v at time t, to v + v at time t + t, we can say that the instantaneous acceleration at time t is a = lim v t 0 t = lim t 0 or, in terms of derivatives a = dv dt = dv x dt, dv y dt v x t, v y t = lim t 0 v x t, lim t 0 v y t (13) In terms of components, a = (a x, a y ) where a x = dv x dt 4and4 a y = dv y dt

39 As usual we can interpret these equations graphically: the value of a x at any particular time is given by the gradient of the v x against t graph at that particular time, and similarly for v y. Express the magnitude of the instantaneous acceleration a in terms of its components a x and a y, and then express the components in terms of the magnitude and the angle ψ from the positive x-axis to a. In what follows we will generally refer to the instantaneous acceleration simply as the acceleration. You learned in the last subsection that one characteristic of projectile motion is that the horizontal component of velocity, v x, is constant. This means that the rate of change of v x must be zero and consequently a x = 0. You also learned that the vertical component of the velocity changes continuously throughout the motion. Careful measurements would show that in the absence of air resistance v y decreases at a constant rate. In other words, throughout the motion the vertical component of velocity is reduced by equal amounts v x in equal intervals of time t, irrespective of when those intervals begin and end. It follows that the vertical component of acceleration is a negative constant which we can write as a y = g. Since this vertical component of acceleration is caused by the action of gravity on the projectile we say that the constant g is the magnitude of the acceleration due to gravity.

40 The value of g varies from place to place over the Earth s surface, but it is generally about 9.81m1s 2 and is usually taken to be 9.811m1s 2 throughout the UK. It describes the acceleration with which all objects near the surface of the Earth fall, so long as they are not impeded appreciably by air resistance. Combining the observations that a x = 0 and a y = g we have: and a = (a x, a y ) = (0, g) (14) a = 1a1 = g It is this particular acceleration that characterizes a projectile near the Earth s surface (in the absence of air resistance). If you were to measure the speed of a projectile at two different times separated by an interval of one second would you always expect them to differ by about 9.81m1s 1? If not, under what conditions would you expect the speeds to differ by about 9.81m1s 1. (As usual, ignore air resistance, but think carefully!)

41 2.5 The independence of x- and y-motions for projectiles As we have seen in Subsections 2.2 and 2.3: The two key features of projectile motion are: 1 A constant horizontal velocity component, v x (since a x = 0). 2 A constant acceleration a of magnitude g = 9.811m1s 2 directed vertically downwards, i.e. a y = 9.811m1s 2, if upwards is taken as the direction of increasing y. These two features will help you to solve a vast range of projectile problems provided you keep one other principle in mind: The horizontal and vertical motions of a projectile must have the same duration, but are otherwise independent of one another and can be treated as two separate one-dimensional (linear) motions. The idea that movement in the x-direction is independent of movement in the y-direction may sound simple and obvious, but many students find it somewhat counter-intuitive and are led into making simple mistakes by forgetting it. The following question is one that many who have not been forewarned might get wrong.

42 Suppose you have two identical bullets and you drop one while firing the other horizontally from a high velocity rifle. Which bullet will hit the ground first? (Ignore air resistance, as usual.) 3 Applying the equations of motion 3.1 Horizontal motion In Subsection 2.4 we saw that a projectile does not experience any acceleration in the horizontal direction. So, in terms of our usual Cartesian coordinate system a x = 0 (15) It follows that the x-component of the projectile s velocity, v x, is constant and will therefore be equal to its initial value at the launch point. So, if this initial value of the horizontal velocity component is denoted by u x, we can write v x = u x (16)

43 It follows from this that if the projectile is launched at time t = 0, then at time t the x-component of its displacement from the launch point will be s x = u x t (17) a x = 0 (Eqn 15) v x = u x (Eqn 16) Equations 15, 16 and 17 are called the uniform motion equations. They are introduced elsewhere in FLAP in the context of one-dimensional (linear) motion, but they are also relevant here because of the independence of the horizontal and vertical motions, and the lack of horizontal acceleration.

44 3.2 Vertical motion In Subsection 2.4 we also stressed that the vertical motion of a projectile is determined by the constant vertical acceleration due to gravity. So, in terms of our usual Cartesian coordinate system a y = g (18) Since this implies that the vertical component of velocity v y decreases at a constant rate from its initial value u y we can say that at time t after launch v y = u y gt (19) Now, since v y is decreasing at a constant rate, its average value between the moment of launch (t = 0) when it has the initial value u y and any later time t when it has the final value v y will be 1v y 1 = (u y + v y )/2. It follows that at time t the vertical component of the projectile s displacement from its launch site will be s y = v y t = u y + v y t = u y + u y gt t 2 2 where Equation 19 has been used to eliminate v y in the last step. Thus s y = u y t 1 2 gt 2 (20)

45 We can relate v y to s y by first squaring both sides of Equation 19 v y = u y gt (Eqn 19) to obtain v y2 = u y2 2u y gt + g 2 t 2 = u y2 2g u y t 1 2 gt 2 and then using Equation 20 s y = u y t 1 2 gt 2 (Eqn 20) to replace the expression in brackets by s y. Thus v y2 = u y2 2gs y (21) Equations 19, 20 and 21 (together with Equations 15, 16 and 17) are the main equations used to solve projectile problems.

46 Equations 19, 20 and 21 are in fact a special case (corresponding to a y = g) of the uniform acceleration equations introduced elsewhere in FLAP v y = u y + a y t (22) s y = u y t a yt 2 (23) v y2 = u y2 + 2a y s y (24) These are slightly more general than Equations 19, 20 and 21 since they apply for any constant value of a y, though it should be noted that they do not describe more general situations in which a y varies with time or position. We will return to Equations 22, 23 and 24 later since, together with Equations 15 to 17, they will allow us to extend our treatment of projectile problems to cover any form of two-dimensional motion where there is uniform acceleration along the y-axis and no acceleration along the x-axis. When using any of Equations 15 to 24 it is important to remember that the displacements s x and s y are measured from the position of the body at time t = 0. Question T6 A ball is thrown vertically upwards with a velocity of 201m1s 1. How long will the ball take to reach the highest point before it begins to fall and what is this maximum height above the point of launch? (Assume g = 9.811m1s 2.)4

47 3.3 The trajectory of a projectile The shape of a projectile s trajectory may be specified in a number of ways, for example Equations 17 and 20 express the displacement components s x and s y in terms of the time t since launch1 1thus providing a parametric description of the trajectory. However, a more informative specification may be obtained by expressing s y directly in terms of s x. We can derive this relationship by eliminating t from Equations 17 and 20 as follows. From Equation 17 t = s x u x Substituting this into Equation 20 gives us s y = u y s x u x 1 2 g s x u x 2 This equation can be rearranged to give s y = u y s x g u x 2u s 2 x 2 (25) x

48 Any projectile, launched with speed u at an angle φ to the x-axis, will initially have velocity components u x = u1cos1φ and u y = u1sin1φ, so Equation 25 s y = u y s x g u x 2u s 2 x 2 (Eqn 25) x becomes s y = usin φ u cos φ s x g 2u 2 cos 2 φ s x 2 = s x tan φ Since u and φ are constants we can write Equation 26 in the form g 2u 2 cos 2 φ s x 2 (26) s y = As x Bs x 2 (27) where A and B are constants. Equations 25 and 27 show that s y is a quadratic function of s x, and imply that the graph of s y against s x will have the shape known as a parabola. The projectile trajectories shown in Figures 3, 5 and 8 were all parabolas. A projectile moving near the Earth, under the influence of gravity alone, follows a parabolic trajectory.

49 3.4 The range of a projectile Archers, gunners and cricketers often want to know how to maximize the range of a projectile, i.e. the horizontal distance between the launching and landing points. This can be easily deduced from Equation 25 (or 26) if the launching and landing points are at the same vertical height because under these conditions the final vertical displacement will be s y = 0 and Equation 25 will give 0 = u y s x g u x 2u s 2 x 2 (Eqn 25) x i.e. u y u x s x = g 2u x 2 s x 2 One solution to this equation is s x = 0, which corresponds to the launching of the projectile. However, this mathematical solution is not of much physical interest. The solution we want (corresponding to the landing of the projectile) has s x 0. We can therefore divide both sides of the equation by s x and rearrange to obtain s x = 2u yu x g

50 This value of s x is the horizontal displacement from the point of projection (the launch) to the landing point. The magnitude of this displacement is the range, R. Therefore R = 2u yu x g (28) We are now in a position to consider a problem which is of interest to cricketers or rounders players. At what angle to the horizontal, φ, should a fielder throw a ball to achieve maximum range? u φ R Figure 94The range of a thrown ball. Figure 9 defines the range

51 while Figure 10 shows the effect on the range of varying the angle of projection. If the ball is thrown too steeply it achieves plenty of height but very little range. On the other hand, if it thrown at too shallow an angle, gravity will pull the ball down before it has chance to travel far. There must be some intermediate value of φ at which the maximum range is achieved. To simplify the problem, we will assume that the ball is thrown from ground level at the same speed whatever the angle and that air resistance is negligible. too steep u φ (a) (b) Figure 104Maximizing the range of a projectile. φ u too shallow

52 Suppose the ball is thrown with an initial velocity u at an angle φ to the horizontal, achieving a range R. The initial (and subsequent) horizontal component of velocity is u x = u1cos1φ and the initial vertical component of velocity is u y = u1sin1φ. Substituting these values into Equation 28 R = gives us R = 2u yu x g 2u2 sin φ cos φ g (Eqn 28) (29) Using the trigonometric identity 21sin1φ1cos1φ = sin1(2φ) Since u 2 and g are positive quantities we can write Equation 29 as R = u2 g sin (2φ ) (30)

53 Assuming φ is between 0 and 180, what are the greatest and least values that sin1(2φ) can have, and hence what is the maximum range? We now need to find the angle of projection, φ, which produces the maximum range. What is the value of φ that gives the maximum range? A projectile moving near the Earth, under the influence of gravity alone, achieves maximum horizontal range when launched at 45 to the horizontal. For the flight of many objects, such as would be used in cricket or in putting the shot, these assumptions are fairly good.

54 Question T7 In the 1976 Olympic Games the shot-putting event was won with a throw of m. Assuming that the shot was thrown from ground level at the optimum angle (45 ), calculate the speed at which it was projected, making use of the formula for range. (Assume g = 9.811m1s 2.)4 In the next section you will learn how to solve similar problems in a more fundamental way, without resorting to the range formula.

55 4 Solving projectile problems Using the ideas developed in the previous two sections you should be able to solve almost any projectile problem. It is particularly important to remember that: In projectile motion, the horizontal component of velocity is constant and the vertical component of acceleration is constant. The usual strategy for dealing with projectile problems is as follows: 1 Express the initial velocity u in terms of its horizontal and vertical components u x = u1cos1φ and u y = u1sin1φ. 2 Treat each component of the motion as an example of one-dimensional (linear) motion: one along a horizontal line at a constant velocity, v x, and the other along a vertical line at a constant acceleration g, (assuming vertically upwards to be the direction in which y increases). 3 Link the two component motions together by means of the total time of flight, T, which must be the same for each.

56 4.1 Some examples of projectile motion Before you tackle a projectile problem on your own, here is a worked example. Example 1 A cannon points at 60 above the horizontal and fires a ball from ground level with a muzzle speed of 20.01m1s 1. Find the horizontal range R, the maximum height h and the time of flight T. Neglect air resistance. Solution4Using the strategy described above to solve this problem, we will first find the horizontal and vertical components of the initial velocity, u x and u y, respectively. u x = (20.0 cos160 )1m1s 1 = 10.01m1s 1 and u y = (20.0 sin160 )1m1s 1 = 17.31m1s 1 We can now make use of the two properties of projectile motion.

57 Horizontal motion4the displacement s x is given by Equation 17 s x = u x t (Eqn 17) The horizontal range R can be found from this equation by putting t = T, (where T denotes the time of flight) and taking the modulus. Therefore R = 1u x T1 (31) We can find T by considering the vertical motion. Vertical motion4when the ball lands on the ground at the end of the flight t = T and the final vertical displacement is s y = 0. So, upon substituting s y = 0 and t = T into s y = u y t 1 2 gt 2 (Eqn 20) we have 0 = u y T 1 2 gt 2 = Tu ( y 1 2 gt ) This is a quadratic equation and therefore has two solutions. These are T = 0 and T = 2u y g = m s m s 2 = 3.53s (32)

58 The first solution, T = 0, is the time the ball leaves the cannon; the second solution corresponds to the ball hitting the ground again and this is the time we need to find the range R. Now that we have found the time of flight we can substitute this value for T into Equation 31, R = 1u x T1 (Eqn 31) giving R = 1u x T1 = 101m1s s = 35.31m. When the ball is at its maximum height, s y = h, and the vertical component of velocity is momentarily zero, so v y = 0. We can substitute these values into Equation 21 v y2 = u y2 2gs y (Eqn 21) to give the maximum height 0 = u y2 2gh i.e. h = u y 2 2g = (17.3 m s 1 ) 2 = 15.3m m s 2

59 Now apply this strategy yourself to the next two questions. Question T8 A cannon standing on the top of a cliff, 401m high, fires a cannonball horizontally out to sea with a muzzle speed of 1401m1s 1. How far out to sea does the ball go? (Assume g = 9.811m1s 2.)4 Question T9 A golf ball is hit on level ground so that it leaves the ground with an initial velocity of 40.01m1s 1 at 30 above the horizontal. Find the greatest height reached by the ball, the total time of flight of the ball and the range of the shot. Neglect air resistance and assume g = 9.811m1s 2.4